In triangle ABC,angle B = frac{pi}{4} and ang | vedclass

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(D) Given,$\angle B = \frac{\pi}{4}$,$\angle C = \frac{\pi}{3}$.In $	riangle ABC$,$\angle A = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\p

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$6(\sqrt{3} + 1)$

Vedclass · Answer

(D) Given,$\angle B = \frac{\pi}{4}$,$\angle C = \frac{\pi}{3}$.In $	riangle ABC$,$\angle A = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} = 75^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$.The area of the triangle is $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} (2R \sin B)(2R \sin C) \sin A = 2R^2 \sin A \sin B \sin C$.Given $\Delta = 54 + 18\sqrt{3} = 18(3 + \sqrt{3})$.$\sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.$\Delta = 2R^2 \left(\frac{\sqrt{6} + \sqrt{2}}{4}ight) \left(\frac{1}{\sqrt{2}}ight) \left(\frac{\sqrt{3}}{2}ight) = 2R^2 \left(\frac{\sqrt{3} + 1}{4\sqrt{2}}ight) \left(\frac{\sqrt{3}}{\sqrt{2}}ight) = 2R^2 \frac{3 + \sqrt{3}}{8} = R^2 \frac{3 + \sqrt{3}}{4}$.Equating the areas: $R^2 \frac{3 + \sqrt{3}}{4} = 18(3 + \sqrt{3})$ $\Rightarrow R^2 = 72$ $\Rightarrow R = 6\sqrt{2}$.Finally,$a = 2R \sin A = 2(6\sqrt{2}) \left(\frac{\sqrt{6} + \sqrt{2}}{4}ight) = 12\sqrt{2} \left(\frac{\sqrt{2}(\sqrt{3} + 1)}{4}ight) = 3(2)(\sqrt{3} + 1) = 6(\sqrt{3} + 1)$.

In $\triangle ABC$,$\angle B = \frac{\pi}{4}$ and $\angle C = \frac{\pi}{3}$. If the area of the triangle is $54 + 18\sqrt{3}$ sq. units,then $a =$

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